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🎲 Chapter 7: The Mathematics of Maybe: Introduction to Probability

Verified step-by-step solutions for your class notes.

Q1

Rank the following events on a scale from 0 (Impossible) to 1 (Certain). Label each event: Impossible, less likely, equally likely (even chance), more likely, certain. Give reasons why you gave each event its ranking.

  • (i) The next Monday will come after Sunday.
  • (ii) It will snow in Mumbai in July.
  • (iii) An elephant will walk through your classroom today.
  • (iv) You will greet at least one friend at school tomorrow.
  • (i) Certain (1): The days of the week are defined in a fixed order, so Monday always follows Sunday.
  • (ii) Impossible (0): Mumbai has a tropical climate and it is currently monsoon season in July; it never snows there.
  • (iii) Impossible (0): Schools are not habitats for elephants, making this scenario practically impossible.
  • (iv) More likely (closer to 1): Assuming you attend school regularly and have friends, it is highly probable you will greet one, though not absolutely guaranteed (e.g., you might be absent).
Q1

A teacher mixes a large bag of sweets of different colours and randomly selects a sample of 30 sweets. She counts: 10 red | 8 green | 7 yellow | 5 blue

  • (i) Calculate the probability that a randomly picked sweet from the sample is green.
  • (ii) If there are 600 sweets in total in the large bag, estimate how many are likely to be yellow, based on the sample results.
  • (i) Probability of green = Number of green / Total in sample = $8 / 30 = \mathbf{4/15}$ (or $\approx 0.267$).
  • (ii) Experimental probability of yellow = $7 / 30$.
    Estimated yellow sweets in total bag = $(7 / 30) \times 600 = 7 \times 20 = \mathbf{140}$ sweets.
Q2

A survey is conducted at a school where a random sample of 40 students is asked about their favourite club. The responses are: 14 Science | 11 Arts | 9 Sports | 6 Debate. Assume there are 800 students in the whole school.

  • (i) What is the probability that a randomly chosen student from the sample prefers the Arts Club?
  • (ii) Using the sample results, estimate how many students in the whole school are likely to prefer the Sports Club.
  • (i) Probability of Arts Club = $11 / 40$ (or $0.275$).
  • (ii) Experimental probability of Sports Club = $9 / 40$.
    Estimated students preferring Sports Club = $(9 / 40) \times 800 = 9 \times 20 = \mathbf{180}$ students.
Q5

What is the probability of getting an even number when rolling a fair 6-sided die?

Answer: The possible outcomes are $\{1, 2, 3, 4, 5, 6\}$.
The even numbers are $\{2, 4, 6\}$.
Probability = Number of favourable outcomes / Total possible outcomes = $3 / 6 = \mathbf{1/2}$ (or $0.5$).

Q6

Suppose you roll a 6-sided die 12 times and get a '3' three times.

  • (i) What is the experimental probability of rolling a '3'?
  • (ii) What is the theoretical probability of rolling a '3'?
  • (iii) Why might these probabilities be different? What would you expect to happen if you roll the die 60, 600, or 6000 times?
  • (i) Experimental probability = $3 / 12 = \mathbf{1/4}$ (or $0.25$).
  • (ii) Theoretical probability = $\mathbf{1/6}$ (or $\approx 0.167$).
  • (iii) They are different because experimental probability is based on a small number of actual trials, which are subject to randomness and short-term fluctuations. If you roll the die 60, 600, or 6000 times, the Law of Large Numbers states that the experimental probability will gradually get closer and closer to the theoretical probability of $1/6$.
Q1

When a single 6-sided die is rolled, what is the total number of possible outcomes in the sample space?

Answer: The total number of possible outcomes (the sample size) is 6. The sample space is $\{1, 2, 3, 4, 5, 6\}$.

Q2

For the following experiments write down the sample space S.

  • (i) Rolling a die and tossing a coin together.
  • (ii) Choosing a random integer between -5 and +5.
  • (iii) A box containing 5 green and 7 red balls. One ball is drawn at random.
  • (i) $S = \{1H, 1T, 2H, 2T, 3H, 3T, 4H, 4T, 5H, 5T, 6H, 6T\}$
  • (ii) Assuming inclusive: $S = \{ -4, -3, -2, -1, 0, 1, 2, 3, 4 \}$.
  • (iii) $S = \{\text{Green}, \text{Red}\}$.
Q3

In a village fair, there are 3 popular snacks available: Samosa, Pakora, and Bhaji. For drinks, villagers can choose either Chai or Lassi.

  • (i) List the sample space of all possible snack and drink combinations a person could choose at the fair.
  • (ii) List the event 'Selecting Samosa as a snack.'
  • (i) $S = \{\text{(Samosa, Chai)}, \text{(Samosa, Lassi)}, \text{(Pakora, Chai)}, \text{(Pakora, Lassi)}, \text{(Bhaji, Chai)}, \text{(Bhaji, Lassi)}\}$
  • (ii) Event $E = \{\text{(Samosa, Chai)}, \text{(Samosa, Lassi)}\}$
Q1

There are two fruit baskets A and B. Basket A has one apple and two oranges. Basket B has one banana and one mango. You randomly pick one fruit from each basket.

  • (i) Draw a tree diagram showing all possible pairs of fruits.
  • (ii) List the sample space.
  • (iii) What is the probability of picking one apple and one banana?
  • (i) Tree Diagram:
    From start, 2 branches for Basket A: Apple (prob 1/3) and Orange (prob 2/3).
    From Apple, 2 branches for Basket B: Banana (prob 1/2) and Mango (prob 1/2).
    From Orange, 2 branches for Basket B: Banana (prob 1/2) and Mango (prob 1/2).
  • (ii) Sample Space: $S = \{\text{(Apple, Banana)}, \text{(Apple, Mango)}, \text{(Orange, Banana)}, \text{(Orange, Mango)}\}$
  • (iii) Probability: P(Apple from A) $\times$ P(Banana from B) = $(1/3) \times (1/2) = \mathbf{1/6}$.
Basket A Basket B Outcome (Prob) 1/3 2/3 Apple Orange 1/2 1/2 Banana Mango 1/2 1/2 Banana Mango (A, B) → 1/6 (A, M) → 1/6 (O, B) → 2/6 = 1/3 (O, M) → 2/6 = 1/3
Tree Diagram of picking one fruit from Basket A and one from Basket B
Q1

Fill in the blanks.

  • (i) The probability of an impossible event is ________________.
  • (ii) The set of all possible outcomes of a random experiment is called the ________________.
  • (iii) The probability of an event that is certain to happen is ________________.
  • (iv) Tossing a fair coin has a probability of ________________ for getting heads.
  • (i) 0
  • (ii) sample space
  • (iii) 1
  • (iv) 1/2 (or 0.5)
Q4

Write the sample space and calculate the probability based on the given information.

  • (i) Two coins are tossed at the same time. What is the probability of getting at least one head?
  • (ii) Ten identical cards numbered 1 to 10 are placed in a box. One card is drawn at random. What is the probability of drawing a card with an even number?
  • (iii) A die is rolled once. What is the probability of getting a number greater than 4?
  • (iv) A bag contains 3 red balls, 2 blue balls, and 1 green ball. One ball is picked at random. What is the probability that it is not red?
  • (v) Three coins are tossed simultaneously. What is the probability of getting exactly two heads?
  • (i) $S = \{HH, HT, TH, TT\}$. Favourable = $\{HH, HT, TH\}$. Probability = $\mathbf{3/4}$.
  • (ii) $S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$. Favourable (even) = $\{2, 4, 6, 8, 10\}$. Probability = $5/10 = \mathbf{1/2}$.
  • (iii) $S = \{1, 2, 3, 4, 5, 6\}$. Favourable (>4) = $\{5, 6\}$. Probability = $2/6 = \mathbf{1/3}$.
  • (iv) $S = \{\text{Red, Red, Red, Blue, Blue, Green}\}$. Favourable (not red) = 3 (2 blue + 1 green). Probability = $\mathbf{3/6 = 1/2}$.
  • (v) $S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}$. Favourable (exactly two heads) = $\{HHT, HTH, THH\}$. Probability = $\mathbf{3/8}$.
Q6

A child has 2 shirts (one red and one blue) and 3 types of pants (jeans, khakis, and shorts). List all the possible combinations of outfits consisting of one shirt and one pair of pants. Display your answer in a table format.

ShirtPantsOutfit Combination
RedJeansRed & Jeans
RedKhakisRed & Khakis
RedShortsRed & Shorts
BlueJeansBlue & Jeans
BlueKhakisBlue & Khakis
BlueShortsBlue & Shorts
Q7

A tyre company records distances before replacement in 1000 cases.

Distance (km)Less than 40004001 to 90009001 to 14000More than 14000
Number of cases20210325445

Find the probability that a randomly chosen tyre lasts:
(i) Less than 4000 km.
(ii) Between 4000 and 14000 km.
(iii) More than 14000 km.

  • (i) Probability = $20 / 1000 = \mathbf{0.02}$ (or 2%).
  • (ii) Favourable cases = $210 + 325 = 535$. Probability = $535 / 1000 = \mathbf{0.535}$ (or 53.5%).
  • (iii) Probability = $445 / 1000 = \mathbf{0.445}$ (or 44.5%).
Q8

The letters of the word 'PEACE' are placed on cards. Leela draws a card without looking.

  • (i) What is the probability that it is a P, E or C?
  • (ii) What is the probability that it is not an E?
  • (i) Total letters = 5. Favourable (P, E, or C) = 4 (P, E, E, C). Probability = $\mathbf{4/5}$.
  • (ii) Favourable (not E) = 3 (P, A, C). Probability = $\mathbf{3/5}$.
Q9

A game of chance consists of spinning an arrow which comes to rest pointing at one of the numbers 1, 2, 3, 4, 5, 6, 7, 8, and these are equally likely outcomes. What is the probability that it will point at

  • (i) 8?
  • (ii) An odd number?
  • (iii) A number greater than 2?
  • (iv) A number less than 9?
  • (v) A multiple of 3?
1 2 3 4 5 6 7 8
8-Sector Equally Likely Spinner Wheel
  • (i) Favourable = 1. Probability = $\mathbf{1/8}$.
  • (ii) Odd numbers = $\{1, 3, 5, 7\}$ (4 outcomes). Probability = $4/8 = \mathbf{1/2}$.
  • (iii) Numbers > 2 = $\{3, 4, 5, 6, 7, 8\}$ (6 outcomes). Probability = $6/8 = \mathbf{3/4}$.
  • (iv) Numbers < 9 = All 8 numbers. Probability = $8/8 = \mathbf{1}$ (Certain).
  • (v) Multiples of 3 = $\{3, 6\}$ (2 outcomes). Probability = $2/8 = \mathbf{1/4}$.
Q10

A basket contains 4 red balls and 5 blue balls. One ball is drawn and laid aside, and a second ball is drawn. Draw a tree diagram to represent the possible outcomes and probabilities. Use the tree diagram to answer the following questions.

  • (i) What is the probability of drawing a red ball and then a blue ball?
  • (ii) What is the probability of drawing 2 blue balls?

Answer:
Tree Diagram Construction:
First draw: P(Red) = 4/9, P(Blue) = 5/9.
Second draw (without replacement, total is now 8 balls):
- If Red was drawn first: P(Red) = 3/8, P(Blue) = 5/8.
- If Blue was drawn first: P(Red) = 4/8, P(Blue) = 4/8.

(i) P(Red then Blue) = P(Red first) $\times$ P(Blue second | Red first) = $(4/9) \times (5/8) = 20/72 = \mathbf{5/18}$.
(ii) P(Blue then Blue) = P(Blue first) $\times$ P(Blue second | Blue first) = $(5/9) \times (4/8) = 20/72 = \mathbf{5/18}$.

1st Draw (9 balls) 2nd Draw (8 balls) Outcome & Prob 4/9 5/9 Red₁ Blue₁ 3/8 5/8 Red₂ Blue₂ 4/8 4/8 Red₂ Blue₂ RR: 12/72 = 1/6 RB: 20/72 = 5/18 BR: 20/72 = 5/18 BB: 20/72 = 5/18
Tree Diagram for two draws without replacement (4 Red, 5 Blue balls)
Q15

Three coins are tossed, and the number of heads is recorded. Which of the following lists is a sample space for this experiment? Why do the other lists fail to qualify as a sample space?

  • (i) {1, 2, 3}
  • (ii) {0, 1, 2}
  • (iii) {0, 1, 2, 3, 4}
  • (iv) {0, 1, 2, 3}

Answer: The correct list is (iv) {0, 1, 2, 3}. When tossing three coins, you can get 0 heads (TTT), 1 head (HTT, THT, TTH), 2 heads (HHT, HTH, THH), or 3 heads (HHH).
The other lists fail because:
(i) It omits the possibility of getting 0 heads.
(ii) It omits the possibility of getting 3 heads.
(iii) It includes 4 heads, which is impossible when only tossing three coins.